2.1 - Hyperbolic identities

2.1 - Hyperbolic identities

Hyperbolic functions satisfy identities resembling trigonometric identities, but a change of sign matters. You will prove the fundamental relation, derive addition and multiple-angle formulae, and choose identities to simplify expressions or solve equations without losing domain restrictions.

The fundamental identity

Why do cosh⁡(ln⁡2)=5/4\cosh(\ln2)=5/4 and sinh⁡(ln⁡2)=3/4\sinh(\ln2)=3/4 give a difference of squares equal to 11? Their squares give 25/16−9/16=125/16-9/16=1. The exponential definitions explain why the same result holds for every real argument.

Adding and subtracting those definitions gives

cosh⁡x+sinh⁡x=ex,cosh⁡x−sinh⁡x=e−x.\cosh x+\sinh x=e^x, \qquad \cosh x-\sinh x=e^{-x}.

Now factor the difference of squares:

cosh⁡2x−sinh⁡2x=(cosh⁡x+sinh⁡x)(cosh⁡x−sinh⁡x)=exe−x=1.\begin{aligned} \cosh^2x-\sinh^2x &=(\cosh x+\sinh x)(\cosh x-\sinh x) \\ &=e^xe^{-x}=1. \end{aligned}

The product is always 11, however large the two squared terms become.

Fundamental hyperbolic identity

For every real xx,

cosh⁡2x−sinh⁡2x≡1.\cosh^2x-\sinh^2x\equiv1.

In the formula booklet.

Dividing by cosh⁡2x\cosh^2x is always valid, because cosh⁡x≥1\cosh x\ge1. It produces 1−tanh⁡2x=sech⁡2x1-\tanh^2x=\operatorname{sech}^2x. Dividing instead by sinh⁡2x\sinh^2x requires x≠0x\ne0, and gives coth⁡2x−cosech⁡2x=1\coth^2x-\operatorname{cosech}^2x=1.

Reciprocal forms

1−tanh⁡2x=sech⁡2x(x∈R),1-\tanh^2x=\operatorname{sech}^2x \quad(x\in\mathbb R), coth⁡2x−cosech⁡2x=1(x≠0).\coth^2x-\operatorname{cosech}^2x=1 \quad(x\ne0).

These two forms are not printed in the formula booklet: derive them from the fundamental identity.

The identity gives magnitudes. The graphs determine which sign to choose: cosh is always positive, whereas sinh and tanh have the sign of their argument.

Worked example 1: recovering cosh from sinh

Question. Given that sinh⁡x=−7/24\sinh x=-7/24, find the exact values of cosh⁡x\cosh x and tanh⁡x\tanh x.

Plan. Use the difference of squares, select the positive cosh value, then form the ratio.

cosh⁡2x=1+(−724)2=625576.\cosh^2x=1+\left(-\frac7{24}\right)^2 =\frac{625}{576}.

Since cosh⁡x>0\cosh x>0,

cosh⁡x=2524,tanh⁡x=−7/2425/24=−725.\cosh x=\frac{25}{24}, \qquad \tanh x=\frac{-7/24}{25/24}=-\frac7{25}.

Answer. cosh⁡x=25/24\cosh x=25/24 and tanh⁡x=−7/25\tanh x=-7/25.

Check. The relation ex=cosh⁡x+sinh⁡xe^x=\cosh x+\sinh x gives ex=3/4e^x=3/4, hence e−x=4/3e^{-x}=4/3. The definitions give (3/4−4/3)/2=−7/24(3/4-4/3)/2=-7/24 and (3/4+4/3)/2=25/24(3/4+4/3)/2=25/24, independently confirming the signs and values.

Keep the same argument but change which function is given. Recovering sinh now requires information about the sign of the argument.

Worked example 2: selecting the negative sinh value

Question. Given that cosh⁡x=25/24\cosh x=25/24 and x<0x<0, find the exact values of sinh⁡x\sinh x and tanh⁡x\tanh x.

Plan. Find sinh⁡2x\sinh^2x, then use x<0x<0 to choose the square root.

sinh⁡2x=(2524)2−1=49576.\sinh^2x=\left(\frac{25}{24}\right)^2-1 =\frac{49}{576}.

Sinh is negative for a negative argument, so

sinh⁡x=−724,tanh⁡x=−725.\sinh x=-\frac7{24}, \qquad \tanh x=-\frac7{25}.

Answer. sinh⁡x=−7/24\sinh x=-7/24 and tanh⁡x=−7/25\tanh x=-7/25.

Check. Since sech⁡x=24/25\operatorname{sech}x=24/25, the separately derived identity gives tanh⁡2x=1−(24/25)2=49/625\tanh^2x=1-(24/25)^2=49/625. Its negative square root is −7/25-7/25, consistent with x<0x<0.

The opposite argument has the same cosh value but opposite sinh and tanh values. Thus a cosh value alone cannot distinguish the two arguments.

Watch out

The trigonometric identity has a plus sign, but cosh⁡2x+sinh⁡2x=1\cosh^2x+\sinh^2x=1 is false in general. The correct relation is cosh⁡2x−sinh⁡2x=1\cosh^2x-\sinh^2x=1; it implies cosh⁡x≥1\cosh x\ge1, not cosh⁡x≤1\cosh x\le1.

Choose the sign from the stated interval in this check.

Addition and double angles

An addition identity lets you combine two arguments without first finding either argument. To derive one, expand the expression on the right using exponentials:

sinh⁡acosh⁡b+cosh⁡asinh⁡b=(ea−e−a)(eb+e−b)+(ea+e−a)(eb−e−b)4=2ea+b−2e−(a+b)4=sinh⁡(a+b).\begin{aligned} &\sinh a\cosh b+\cosh a\sinh b \\ &=\frac{(e^a-e^{-a})(e^b+e^{-b}) +(e^a+e^{-a})(e^b-e^{-b})}{4} \\ &=\frac{2e^{a+b}-2e^{-(a+b)}}4 \\ &=\sinh(a+b). \end{aligned}

The mixed exponential terms cancel. Expanding cosh⁡acosh⁡b+sinh⁡asinh⁡b\cosh a\cosh b+\sinh a\sinh b in the same way gives (ea+b+e−(a+b))/2(e^{a+b}+e^{-(a+b)})/2, so it equals cosh⁡(a+b)\cosh(a+b).

Addition and subtraction

For real a,ba,b,

sinh⁡(a±b)=sinh⁡acosh⁡b±cosh⁡asinh⁡b,\sinh(a\pm b)=\sinh a\cosh b\pm\cosh a\sinh b, cosh⁡(a±b)=cosh⁡acosh⁡b±sinh⁡asinh⁡b.\cosh(a\pm b)=\cosh a\cosh b\pm\sinh a\sinh b.

The subtraction forms follow by replacing bb by −b-b and using parity. These formulae are not printed in the formula booklet: learn them or derive them from exponentials.

Dividing the sinh addition formula by the cosh addition formula, then dividing top and bottom by cosh⁡acosh⁡b\cosh a\cosh b, also gives

tanh⁡(a±b)=tanh⁡a±tanh⁡b1±tanh⁡atanh⁡b.\tanh(a\pm b) =\frac{\tanh a\pm\tanh b}{1\pm\tanh a\tanh b}.

For real finite arguments ∣tanh⁡atanh⁡b∣<1|\tanh a\tanh b|<1, so these denominators are nonzero. This tanh formula is not printed in the booklet either.

Worked example 3: adding two arguments

Question. Given a=ln⁡2a=\ln2 and b=ln⁡3b=\ln3, use addition identities to find sinh⁡(a+b)\sinh(a+b) and cosh⁡(a+b)\cosh(a+b) exactly.

Plan. Evaluate sinh and cosh at the separate arguments, then substitute in the two addition formulae.

The definitions give

sinh⁡a=34,cosh⁡a=54,sinh⁡b=43,cosh⁡b=53.\sinh a=\frac34,\quad\cosh a=\frac54, \qquad \sinh b=\frac43,\quad\cosh b=\frac53.

Therefore

sinh⁡(a+b)=3453+5443=3512cosh⁡(a+b)=5453+3443=3712.\begin{aligned} \sinh(a+b)&=\frac34\frac53+\frac54\frac43 =\frac{35}{12} \\ \cosh(a+b)&=\frac54\frac53+\frac34\frac43 =\frac{37}{12}. \end{aligned}

Answer. sinh⁡(a+b)=35/12\sinh(a+b)=35/12 and cosh⁡(a+b)=37/12\cosh(a+b)=37/12.

Check. Logarithm laws give a+b=ln⁡6a+b=\ln6. The exponential definitions give (6−1/6)/2=35/12(6-1/6)/2=35/12 and (6+1/6)/2=37/12(6+1/6)/2=37/12, without using addition identities.

Change only the addition to subtraction. The terms stay the same, but the second product now has a minus sign in each formula.

Worked example 4: subtracting the same arguments

Question. Given a=ln⁡2a=\ln2 and b=ln⁡3b=\ln3, use subtraction identities to find sinh⁡(a−b)\sinh(a-b) and cosh⁡(a−b)\cosh(a-b) exactly.

Plan. Reuse the four separate values, subtracting the second product in each formula.

sinh⁡(a−b)=3453−5443=−512cosh⁡(a−b)=5453−3443=1312.\begin{aligned} \sinh(a-b)&=\frac34\frac53-\frac54\frac43 =-\frac5{12} \\ \cosh(a-b)&=\frac54\frac53-\frac34\frac43 =\frac{13}{12}. \end{aligned}

Answer. sinh⁡(a−b)=−5/12\sinh(a-b)=-5/12 and cosh⁡(a−b)=13/12\cosh(a-b)=13/12.

Check. Here ea−b=2/3e^{a-b}=2/3 and e−(a−b)=3/2e^{-(a-b)}=3/2. Their half-difference is −5/12-5/12 and their half-sum is 13/1213/12. The negative sinh value also agrees with a−b<0a-b<0.

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